When anhydrous iron sulfate decomposes it forms iron (III) oxide, sulfur (IV) dioxide and sulfer (VI) oxide. Write a balanced equation for this reaction.
My answer is this: FeSO4 ----> Fe2O3 + SO2 + SO3
Is that right?
To determine if your answer is correct, let's go through the process of balancing the equation step by step:
Step 1: Write the unbalanced equation:
FeSO4 → Fe2O3 + SO2 + SO3
Step 2: Count the number of atoms on each side of the equation:
On the left side of the equation, there is 1 Fe, 1 S, and 4 O atoms.
On the right side of the equation, there are 2 Fe, 3 S, and 8 O atoms.
Step 3: Balance the Fe atoms:
Since there are 2 Fe atoms on the right side, you need to add a coefficient of 2 in front of FeSO4 on the left side:
2 FeSO4 → Fe2O3 + SO2 + SO3
Step 4: Balance the S atoms:
There are 1 S atom on the left side, but there are 3 S atoms on the right side. To balance the S atoms, add a coefficient of 3 in front of SO2 on the left side:
2 FeSO4 → Fe2O3 + 3 SO2 + SO3
Step 5: Balance the O atoms:
There are 4 O atoms on the left side, but there are 11 O atoms on the right side. To balance the O atoms, add a coefficient of 6 in front of SO3 on the left side:
2 FeSO4 → Fe2O3 + 3 SO2 + 6 SO3
Now, let's check if the atoms are balanced on both sides of the equation:
On the left side: 2 Fe x 1 + 2 S x 1 + 12 O x 4 = 2 Fe + 2 S + 48 O
On the right side: 1 Fe x 2 + 3 S x 1 + 9 O x 2 = 2 Fe + 3 S + 18 O
The number of atoms is now balanced on both sides of the equation, so the balanced equation for the decomposition of anhydrous iron sulfate is:
2 FeSO4 → Fe2O3 + 3 SO2 + 6 SO3
Therefore, your initial answer is not correct, and the balanced equation is 2 FeSO4 → Fe2O3 + 3 SO2 + 6 SO3.